Tag: introduction to work

Questions Related to introduction to work

The correct relation between joule and erg is:

physics energy and its forms introduction to work work introduction to work and energy

  1. $1\ J = 10^{-5} erg$

  2. $1\ J = 10^{5} erg$

  3. $1\ J = 10^{-7} erg$

  4. $1\ J = 10^{7} erg$

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Correct Option: D
Explanation:

Joule and erg both are units of work done. An erg is the amount of work done by applying a force of one dyne for a distance of one centimeter. In the CGS base units, it will be one gram centimeter-squared per second-squared. Whereas joule is the amount of work done by applying a force of one newton for a distance of one meter.

Thus,

$1 joule = 1 newton \times 1 m\\$

$1 joule = \dfrac{1 kg \times 1 m}{1 s ^{2}} \times 1 m\\$

$1 J = \dfrac{1000 g \times 100 cm}{1 s ^{2}} \times 100 cm \\$

$1 J = 10^{7} \ erg$

Thus option D is correct.

State the wrong statement 

physics energy and its forms introduction to work work introduction to work and energy

  1. Total work done by internal force in a system is always zero

  2. Work done is different as seen from different frames of reference

  3. In the absence of external forces and non-conservation force , the molecules energy of a system remains conserved

  4. a non conservation force always  do negative work

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Correct Option: D

A force $\vec {F} = -k(x\hat {i} + y\hat {j})$, where $k$ is positive constant, acts on a particle moving in the $x-y$ plane. Starting from the origin, the particle is taken along the positive x-axis to the point $(a, 0)$ and then parallel to the y-axis to the point $(a, a)$.

physics energy and its forms introduction to work work introduction to work and energy

  1. Work done by the force in moving particle along x-axis is $-\dfrac {1}{2}ka^{2}$

  2. Work done by the force in moving particle along x-axis is $-ka^{2}$

  3. Work done by the force in moving particle along y-axis is $-\dfrac {1}{2}ka^{2}$

  4. Total work done by the force for overall motion is $-ka^{2}$

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Correct Option: B,C

A man carries a load on his head through a distance of 5 m. The maximum amount of work is done when he

physics energy and its forms introduction to work work introduction to work and energy

  1. Movies it over an inclined plane

  2. Movies it over a horizontal surface

  3. Lift it vertically upwards

  4. All of the above

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Correct Option: C
Explanation:

The maximum work done by man will be when he lift it vertically upwards because in such situation the man has to exert force opposite to gravity that is in the direction of the displacement of load. 

A force of $5 N$ is applied on a $20 kg$ mass at rest. the work done in the third second is:-

physics energy and its forms introduction to work work introduction to work and energy

  1. $\dfrac{25}{8}J$

  2. $\dfrac{25}{4}J$

  3. $12 J$

  4. $25 J$

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Correct Option: A
Explanation:

Displacement in third second = displacement till 3rd second - Displacement till 2nd second 

=$\dfrac{a}{2}3^2-\dfrac{a}{2}2^2=\dfrac{5}{8}m$ (No term of ut because u=0)

Thus work done in third second =$5\times\dfrac{5}{8}=\dfrac{25}{8}J$

A small ball bearing is releases at the top of a long vertical column of glycerine of height $2h$. The ball bearing falls through a height $h$ in a time $t _{1}$ and then the remaining height with the terminal velocity in time $t _{2}$ Let $W _{1}$ and $W _{2}$ be the work done against viscous drag over these height. therefore.

physics energy and its forms introduction to work work introduction to work and energy

  1. $t _{1}< t _{2}$

  2. $t _{1}> t _{2}$

  3. $W _{1}=W _{2}$

  4. $W _{1}< W _{2}$

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Correct Option: A

No work is done by a force on an object if 

physics energy and its forms introduction to work work introduction to work and energy

  1. the force is always perpendicular to its velocity

  2. the force is always perpendicular to its acceleration

  3. the object is stationary but the point of application of the force moves on the object

  4. the object moves in such a way that the point of application of the force remains fixed.

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Correct Option: A
Explanation:
The work done by a force F in displacing a particle through d is:
$W=\vec{F}\cdot \vec{d}=Fd\cos \theta$
If $\theta =90^o$
$W=Fd\cos 90^o=0$
Now direction of displacement is same as the direction of velocity.
So if force is always perpendicular to the velocity then no work is done.

A moving particle is acted upon by several forces $F _1, F _2, F _3 .....$ etc. One of the force is chosen, say $F _2$,  then which of the following statement about $F _2$ will be true.

physics energy and its forms introduction to work work introduction to work and energy

  1. Work done by $F _2$ will be negative if speed of the particle decreases

  2. Work done by $F _2$ will be positive if speed of the particle increases

  3. Work done by $F _2$ will be equal to the work done by other forces if speed of the particle does not change

  4. If $F _2$ is a conservative force, then work done by all other forces will be equal to change in potential energy due to force $F _2$ when speed remains constant.

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Correct Option: A

The unit kg m$^2$s$^{-2}$ is associated with :

physics energy and its forms introduction to work work introduction to work and energy

  1. work only

  2. kinetic energy only

  3. potential energy only

  4. all the above

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Correct Option: D
Explanation:

The given unit is that of Work and Energy.


Hence all the physical quantities mentioned have the same unit.

Option D is correct
−2

The unit N-s is equivalent to:

physics energy and its forms introduction to work work introduction to work and energy

  1. J

  2. kgms$^{-1}$

  3. kgms$^{-2}$

  4. kgms

Show answer
Correct Option: B
Explanation:

unit of force in Newton($N$).

force =mass$\times$acceleration.
in units $N=kg.m.sec^{-2}$
so $N.s=kg.m.sec{-1}$
so best possible answer is option B.